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Suppose $\sum a_n$ is a series with decreasing, nonnegative terms. Define $$s_n = \sum_{k=1}^n a_k = a_1+\cdots +a_n $$ and $$t_n = \sum _{k=0}^n 2^ka_{2^k} = a_1+2a_2+4a_4+\cdots +2^na_{2^n}.$$

Also, $s_n$ is defined for all $n\geq 1$ whereas $t_n$ is defined for all $n\geq 0$. I need to prove that $s_{2^{n+1}-1}\leq t_n$ for all $n\geq 0$.

My thinking is to use the fact that $a_n \geq 0$ and that $a_n$ is decreasing. How can I use grouping to compare and make this proof?

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    thanks for the edit everyone, any idea how to work through this proof? – 2017-01-20
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    Hey, could you remind me of the proof of how $\sum \frac{1}{n}$ doesn't converge? – 2017-01-20
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    harmonic series – 2017-01-20
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    Take a look at the proof for why the Harmonic series does not converge. – 2017-01-20

2 Answers 2

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Use induction.

$$s_{2^{n+1} - 1} = s_{2^n - 1} + \sum_{k=2^n}^{2^{n+1} - 1} a_k \le t_{n-1} + (2^{n+1} - 1 - 2^n + 1)a_{2^{n}} = t_n $$

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    does this show $s_{2^{n+1}-1}\leq t_n$ for all $n\geq 0$. – 2017-01-20
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HINT Compare $2a_2$ to $a_2 + a_3.$ Compare $4a_4$ to $a_4+a_5+a_6+a_7,$ etc.

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    so $2a_2>a_2+a_3$ and $2^na_2^{n}>a_2^{n}+...a_2^{n+1}-1$ ?? – 2017-01-20
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    @hburt Yeah, that's the idea, I think, modulo some tex rendering issues – 2017-01-20
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    but from here how do I prove the rest? – 2017-01-20
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    @hburt Isn't that kind of the nature of a hint? Write out both expressions and compare $t_k$ to the corresponding $2^{k-1}$ in $a$.. – 2017-01-20
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    $$t_n = \sum _{k=0}^n 2^ka_{2^k} = a_1+2a_2+4a_4+\cdots +2^na_{2^n}.$$ and $$s_n = \sum_{k=1}^n a_k = a_1+\cdots +a_n $$ So since $a_n$ is decreasing then $2a_2>a_2+a_3$ and $4a_4 > a_4+a_5+a_6+a_7$ and so on, so $2^na{_2^n}> a_2^n+...a_2^{n+1}-1$ so therefore $s_{2^{n+1}-1}\leq t_n$ for all $n\geq 0$. – 2017-01-20
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    how does that look? – 2017-01-20
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    @hburt Good. You might want to make sure that the last term lines up right when you write down $s_{2^{n+2}-2}$ and $t_n$ for a given $n$. (Or use Open Ball's induction approach where you only need to compare the last term). – 2017-01-20